hockey puck

[Speags]

a hockey puck of mass m is sliding in the +x direction across a horizontal ice surface. while sliding, the puck is subject to two forces that oppose its motion: a constant sliding friction force of magnitude f, and a air resistance force of magnitude cv^2 , where c is a constant and v is the puck's velocity. At time t=0, the puck's position is x=0, and it's velocity is v_{o} In terms of the given parameters (m,f,c, and v_o), determine:
a) how far the puck slides, that is determine it's position x when it comes to rest;
for a) i got F=-(f+cv^2)
m\frac{dv}{dx}\frac{dx}{dt}=-(f+cv^2)
mvdv=-(f+cv^2)
\frac{mvdv}{(f+cv^2)}=-dx
takeing the intergral of both sides you get (i think)
\frac{m}{2c} ln(f+cv^2)=-x

x=-\frac{m}{2c}ln(f+cv^2)

now the b) part asks how long does the puck slide, that is, determine the time t at which it comes to rest.
i think i need to turn a v into \frac{dx}{dt}
but i'm not sure where to start or how

[HallsofIvy]

You did fine up to this point:
\frac{m}{2c} ln(f+cv^2)=-x
Should be
\frac{m}{2c} ln(f+cv^2)=-x + C
where "C" is a constant of integration. If we take x=0 initially, then
\frac{m}{2c}ln(f+cv_0^2)= C
so you have
\frac{m}{2c} ln(f+cv^2)=-x+ \frac{m}{2c}ln(f+cv_0^2)
and you may want to write that as
\frac{m}{2c}(ln(f+cv^2)-ln(f+cv_0^2))= -x
or
x= \frac{m}{2c}(ln(\frac{f+cv_0^2}{f+cv^2})

Now, what is x when v= 0?

now the b) part asks how long does the puck slide, that is, determine the time t at which it comes to rest.
i think i need to turn a v into \frac{dx}{dt}[/tex]
but i'm not sure where to start or how

The simplest thing to do is go back to your original equation:
m\frac{dv}{dt}= -(f+cv^2)
and don't convert to x. You get
[tex] m\frac{dv}{f+cv^2}= -dt
Can you integrate that? (Think: arctangent.)
Remember that v= v0 when t= 0 and solve for t when v= 0.